# Thread: length of a line

1. ## length of a line

Find the length of the line of $\displaystyle y = \frac {2}{3} x^\frac {3}{2}$ from x=0 to x=2

i used the formula, i dont know if i was supposed to use u substitution, but i got $\displaystyle 2 \sqrt{3}$

2. Originally Posted by >_<SHY_GUY>_<
Find the length of the line of $\displaystyle y = \frac {2}{3} x^\frac {3}{2}$ from x=0 to x=2

i used the formula, i dont know if i was supposed to use u substitution, but i got $\displaystyle 2 \sqrt{3}$
sorry, that is incorrect ... how did you work it?

3. Originally Posted by skeeter
sorry, that is incorrect ... how did you work it?
i admire your help skeeter, i appreciate that you are helping me

well i used the formula and got :

$\displaystyle \int \sqrt{1 + x^\frac {1}{2} dx } from 0 - 2$ i am supposed to take the derivative of the initial equation and use it in the eq. yes?

4. Originally Posted by >_<SHY_GUY>_<
i admire your help skeeter, i appreciate that you are helping me

well i used the formula and got :

$\displaystyle \int \sqrt{1 + x^\frac {1}{2} dx } from 0 - 2$ i am supposed to take the derivative of the initial equation and use it in the eq. yes?
mistake in your "formula". here is the correct version ...

$\displaystyle S = \int_a^b \sqrt{1 + [f'(x)]^2 } \, dx$

the derivative is squared

5. Originally Posted by skeeter
mistake in your "formula". here is the correct version ...

$\displaystyle S = \int_a^b \sqrt{1 + [f'(x)]^2 } \, dx$

the derivative is squared
oh, i know, i squared it on my work here, but i still got the same answer...because i get 1 + x inside the radical, yes?

6. Originally Posted by >_<SHY_GUY>_<
oh, i know, i squared it on my work here, but i still got the same answer...because i get 1 + x inside the radical, yes?
what is your antiderivative and how did you evaluate it?

7. Originally Posted by skeeter
what is your antiderivative and how did you evaluate it?
i dont know what you mean, but from the eq.

$\displaystyle \int_0^2 \sqrt{1 + (x^\frac {1}{2})^2 dx}$
from there i got
$\displaystyle \int_0^2 \sqrt{1+x dx}$

then; $\displaystyle \int_0^2 (1 + x)^ \frac{1}{2}$

and went on from there

8. Originally Posted by >_<SHY_GUY>_<
i dont know what you mean, but from the eq.

$\displaystyle \int_0^2 \sqrt{1 + (x^\frac {1}{2})^2 dx}$
from there i got
$\displaystyle \int_0^2 \sqrt{1+x dx}$

then; $\displaystyle \int_0^2 (1 + x)^ \frac{1}{2}$

and went on from there
"went from there", how?

9. Originally Posted by skeeter
"went from there", how?
after that, i just took the antiderivative and got

$\displaystyle \frac {2}{3} (1 + x)^ \frac {3}{2}$ from 0 - 2
i then substituted the values. is that ok?

10. Originally Posted by >_<SHY_GUY>_<
after that, i just took the antiderivative and got

$\displaystyle \frac {2}{3} (1 + x)^ \frac {3}{2}$ from 0 - 2
i then substituted the values. is that ok?
keep going ... show me how you arrived at the value $\displaystyle 2\sqrt{3}$.

11. Originally Posted by skeeter
keep going ... show me how you arrived at the value $\displaystyle 2\sqrt{3}$.
then it comes up to be $\displaystyle \frac {2}{3} (\sqrt{3})^3$
i end up with a fraction which is simplified to 2rad 3

12. Originally Posted by >_<SHY_GUY>_<
then it comes up to be $\displaystyle \frac {2}{3} (\sqrt{3})^3$
i end up with a fraction which is simplified to 2rad 3
ok ... $\displaystyle \frac{2}{3}(1+2)^{\frac{3}{2}} = 2\sqrt{3}$

but ...

what did you get when you evaluated $\displaystyle \frac{2}{3}(1+x)^{\frac{3}{2}}$ using the lower limit of integration, $\displaystyle x = 0$ ?

... or did you just assume it would have a value of 0 ?

13. Originally Posted by skeeter
ok ... $\displaystyle \frac{2}{3}(1+2)^{\frac{3}{2}} = 2\sqrt{3}$

but ...

what did you get when you evaluated $\displaystyle \frac{2}{3}(1+x)^{\frac{3}{2}}$ using the lower limit of integration, $\displaystyle x = 0$ ?

... or did you just assume it would have a value of 0 ?
thats my mistake isnt it?

now i get it...i worked to fast on the problem and disregarded the 1

thank you